Almer S. Tigelaar

A Little Bit of Everything

Puzzle #1: The Student Admission Problem

Background: I will occasionally present a small (mathematical) puzzle here. These are a tribute to the column “Vuiks Verhandelingen”, written by Kees Vuik which regularly appeared in the Dutch PCM Magazine nearly a decade ago. Sometimes I may think of them myself, and sometimes they may come from elsewhere.

Let’s get started: 114 students applied for a specific study, only 100 were eventually admitted by lottery. Among the 114 students were 5 foreign students. However, none of these foreigners were among the 100 that were admitted. Intuitively this seems quite unlikely. But how unlikely is it really? Were the foreigners intentionally not admitted?

Your task: find the probability that the 5 foreign students were left out due to chance. This can be solved with basic knowledge of probability and set theory. First try it yourself, then click below to expand the solution and check your answer.


Tip: if you are not too familiar with mathematical notation, don’t be intimidated by it: it’s just a way to formally write things down, and the notation here should be easy to follow. I’ve added some footnotes to help: hover over them with your mouse-pointer to see an explanation.


We start with some generic definitions:

  • S is the set of all students that applied.
  • F is the subset of S which is foreign.
  • n is the number of students we may select from S (without replacement1).
  • The resulting selection we call set L (where L is a subset of S with cardinality2 n )

Let’s fill out the variables above for this specific case:

  • There are 114 students in total: #S=114
  • Of which 5 are foreigners: F\subset S\wedge#F=5 3
  • We may select 100 students: n=100


What is the probability that we select no foreign student when we choose 100 students out of the 114 that applied? Formal: what is the probability that the intersection of F and L is empty?: P\left(F\cap L=\varnothing\right)


  1. We know that there are \binom{#S}{n} ways4 in which we can select n students from the #S (without replacement).
  2. There are \binom{#F}{k} ways to select k students from the foreign subset.
  3. Given (2) there are \binom{#S-#F}{n-k} ways to select the remaining n - k students.
  4. Combining (1-3) we get:P\left(F\cap L=\varnothing\right)=\frac{\binom{#F}{k}\cdot\binom{#S-#F}{n-k}}{\binom{#S}{n}}

In this last step we effectively state that the probability is the number of ways in which we can select k and n - k students, relative to the total number of possible selections we can make. It is a generic solution for any set sizes, for this specific instance we take the problem specific values and simply substitute them:

P\left(F\cap L=\varnothing\right)=\frac{\binom{5}{0}\cdot\binom{114-5}{100-0}}{\binom{114}{100}}=\frac{\binom{5}{0}\cdot\binom{109}{100}}{\binom{114}{100}}=\frac{\binom{109}{100}}{\binom{114}{100}}\thick\approx1.36\cdot10^{-5}

All the values are as specified in the problem specific definition. The value of k is 0, since we are interested in the probability of selecting 0 students from the Foreign group. We see that the problem simplifies, since \binom{5}{0}=1.

We can conclude that the probability of selecting none of the 5 foreign students is quite low: 0.00136 %

The probability that the lottery was unfair can be deduced from this: P\left(F\cap L\neq\varnothing\right)=1-P\left(F\cap L=\varnothing\right)=1-1.36\cdot10^{-5}=0.9999864

Which is 99.99864%. Indeed, it is very likely that at least one foreign student would have been selected. Hence we can state that it is likely that the lottery applied was unfair in this case and that the probability of a foreign student of being selected must have been much lower than that of a local student, or even zero.


1) This means that the same student can not be selected twice. So, while each student initially has a chance of 1/114 to be picked, after the first pick this will be 1/113, then 1/112, et cetera.

2) A fancy way of saying: size.

3) From left to right: F is a subset of S and there are 5 students in F.

4) This is binomial notation, that should be interpreted as: \binom{n}{k}=\frac{n!}{k!\cdot\left(n-k\right)!}. With:

  • n!=prod_{r=1}^{n}r=n\cdot\left(n-1\right)\cdot\left(n-2\right)\ldots1
  • 0!=1
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